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Home Impedancematching Networks Type III Problem  
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Type III ProblemAuthor: Edmund A. Laport
Procedure. Write the problem vectorially as shown in Fig. 5.25. Because input impedance equals the load impedance, the input is represented by vectors of the same length as for the load, but with the specified phase difference. Then, if we chose a T network (Fig. 5.26), we erect perpendiculars through V_{0} and V_{in} and connect I_{0} with I_{in} as shown in Fig. 5.27. From this we obtain the solution shown in Fig. 5.28.
Had we chosen the π solution, the vector diagram would have been as shown in Fig. 5.29. A reversal of the phase angle in any problem reverses the signs of all the elements. For instance, in this problem, if the phase angle had been + 120 degrees, all elements would change from L to C, and vice versa, maintaining the same numerical values. A phaseshifting network of this type is equivalent to a section of an infinite transmission line having a characteristic impedance R_{0} and an electrical length equal to that of the angle of phase shift, which in such cases must be negative. This provides a handy device to study certain properties of lines, such as the 45 degree and the 90 degree sections which have special properties.
For example, how does the input impedance of a 45 degree section of a transmission line change with variation of the terminal resistance from 0 to ? Solve for the equivalent singlestage network having a characteristic impedance R_{0} when it is terminated in a resistance R_{0} by assuming it to be a 45 degree phaseshifting network. (We illustrate with the T network.) The vector diagram is shown in Fig. 5.30. Thus, taking R_{0} as unity, we obtain the ratio of all other elements to R_{0} . We can then assume any value of terminal resistance and calculate the resulting input impedance. We obtain the circuit of Fig. 5.31. The limiting conditions are quickly examined by assuming opencircuit and shortcircuit terminal conditions. For the former, Z_{in} is the sum of j0.42 and j1.42, which is j1.00. For the latter, the solution for Z_{in} yields +j1.00. So we easily find that Z_{in} remains constant in magnitude for all conditions of resistance termination but varies in character from pure capacitive reactance, through pure resistance to pure inductive reactance as the terminal resistance varies from to 0.
Proceeding in the same manner, the 90degree network is analyzed.
Its special property of impedance inversion is worthy of digression to see why it works as it does. The vector diagram for a 90degree T network with a characteristic impedance R_{0} is given in Fig. 5.32. From this we find the equivalent circuit elements, in terms of R_{0} , to be those of Fig. 5.33. All the elements of the network have reactances equal in magnitude to R_{0} .
At this point we can test the circuit for open and shortcircuit terminal conditions and see what happens to Z_{in}. When R_{t} = R_{0} , Z_{in} = R_{0} . When R_{t} = 0, X_{1} and X_{2} in parallel give a condition of antiresonance and Z_{in} = . When R_{t} = , X_{2} and X_{3} in series tune to series resonance, making Z_{in} = 0.


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